Triangular Form and Gaussian Elimination
Boldly on to Sec. 7.3a…HW: p. 602 1-19 odd
Triangular Form – of a system of equations has the leadingterm of each equation with coefficient 1, the final equation hasonly one variable, and each higher equation has one additionalvariable
Example: 2 7x y z 2 7y z
3z
Gaussian Elimination – the process of transforming a systemto triangular form. Gaussian elimination will lead to another type of matrices manipulation.
Steps that can be used in Gaussian Elimination (all of whichproduce equivalent systems of linear equations):
1. Interchange any two equations of the system.
2. Multiply (or divide) one of the equations by any nonzero real number.
3. Add a multiple of one equation to any other equation in the system.
4. Replace the equation, and continue the process.
Back to our original example Solve by substitution!
2 7x y z 2 7y z
3z 2 7x y z
2 3 7y 3z
1y
2 1 3 7x 1y 3z
2x Solution:
, , 2, 1,3x y z
Another Example – Solve using Gaussian Elimination:
2 7x y z 3 5 14x y z 2 2 3x y z
Multiply the first equation by –3 andadd the result to the second equation,replacing the second equation
2 7x y z 2 7y z
2 2 3x y z 1 23E E
Another Example – Solve using Gaussian Elimination:
Multiply the first equation by –2 andadd the result to the third equation,replacing the third equation.
2 7x y z 2 7y z
2 3 11y z 1 32E E
2 7x y z 2 7y z
2 2 3x y z
Another Example – Solve using Gaussian Elimination:
Multiply the second equation by –2and add the result to the third equation,replacing the third equation.
2 7x y z 2 7y z
2 3 11y z
2 32E E 2 7x y z 2 7y z
3z This is our first example!!!This is our first example!!!
Solve using Gaussian Elimination:
3 4x y z 2 5 3x y z
5 13 13 8x y z
1 2E E
2 32E E1 35E E
Steps:
3 4x y z 4 7y z 0 2
This last equation is never true…This last equation is never true… No Solution!!!No Solution!!!
Solve using Gaussian Elimination:
Solution: (x, y, z) = (5/13, 10/13, 74/13)Solution: (x, y, z) = (5/13, 10/13, 74/13)
2 0x y 3 3x y z 3 8y z
12E 2 0x y 3 3x y z
3 8y z
1 22E E 7 2 6y z 3 3x y z
3 8y z
2 3
3E E7
7 2 6y z 3 3x y z
13 7 74 7z
Solve using Gaussian Elimination:
0.5 1x y z w 2 3x y z w
2x z 0y w
3 2E E0.5 1x y z w
2 1y w 2x z 0y w
4 2E E 0.5 1x y z w
1w2x z 0y w
3 1
1E E2
1.5 0y z w
1w2x z 0y w
Solution: (x, y, z, w) = (2, 1, 0, –1)Solution: (x, y, z, w) = (2, 1, 0, –1)